Spiral Matrix Traversal in DSA Rust - Build from Scratch

Spiral Matrix Traversal

📖 Scenario: Imagine you have a square garden divided into a grid. You want to walk around the garden starting from the top-left corner, moving right, then down, then left, then up, and keep going in a spiral until you have visited every section.

🎯 Goal: You will write a Rust program that takes a 3x3 matrix of numbers and prints the numbers in the order you would visit them if you walked in a spiral around the matrix.

📋 What You'll Learn

Create a 3x3 matrix with exact values

Set up variables to track the boundaries of the matrix

Use a loop to traverse the matrix in a spiral order

Print the spiral order as a list

💡 Why This Matters

🌍 Real World

Spiral traversal is useful in image processing, game development, and robot path planning where you need to visit elements in a spiral pattern.

💼 Career

Understanding matrix traversal techniques helps in coding interviews and solving problems related to grids and 2D data structures.

1

Create the 3x3 matrix

Create a variable called matrix that is a 3x3 vector of vectors with these exact values: vec![vec![1, 2, 3], vec![4, 5, 6], vec![7, 8, 9]]

DSA Rust

# Create the 3x3 matrix called matrix
# Your code here

Hint

Use vec![vec![...], vec![...], vec![...]] to create a 2D vector.

2

Set up boundary variables

Create four variables called top, bottom, left, and right. Set top to 0, bottom to 2, left to 0, and right to 2.

DSA Rust

let matrix = vec![vec![1, 2, 3], vec![4, 5, 6], vec![7, 8, 9]];
# Set up top, bottom, left, right boundaries
# Your code here

Hint

These variables will help track the edges of the matrix as you move inward.

3

Traverse the matrix in spiral order

Create a mutable vector called result to store the spiral order. Use a while loop that runs while left <= right and top <= bottom. Inside the loop, add the top row elements from left to right to result, then increment top. Next, add the right column elements from top to bottom to result, then decrement right. Then, if top <= bottom, add the bottom row elements from right to left in reverse to result, then decrement bottom. Finally, if left <= right, add the left column elements from bottom to top in reverse to result, then increment left.

DSA Rust

let matrix = vec![vec![1, 2, 3], vec![4, 5, 6], vec![7, 8, 9]];
let top = 0;
let bottom = 2;
let left = 0;
let right = 2;

# Create result vector and write the spiral traversal loop
# Your code here

Hint

Use for loops and push to add elements to result. Use saturating_sub(1) to safely subtract 1 from unsigned integers.

4

Print the spiral order

Print the result vector using println!("{:?}", result) to show the spiral traversal order.

DSA Rust

Hint

Use println!("{:?}", result) to print the vector in Rust.

Practice

1. What is the main idea behind Spiral Matrix Traversal?

easy

A. Traverse the matrix column by column from left to right

B. Traverse the matrix row by row from top to bottom

C. Traverse the matrix in a circular pattern from outer to inner layers

D. Traverse only the diagonal elements of the matrix

Solution

1. Step 1: Understand traversal pattern

   Spiral traversal means moving around the edges of the matrix and gradually moving inward in a circular way.

2. Step 2: Compare options

   Only Traverse the matrix in a circular pattern from outer to inner layers describes this circular, layer-by-layer movement. Others describe linear or partial traversals.

3. Final Answer:

   Traverse the matrix in a circular pattern from outer to inner layers -> Option C

4. Quick Check:

   Spiral traversal = circular outer to inner [OK]

Hint: Think of peeling an onion layer by layer [OK]

Common Mistakes:

• Confusing spiral with row-wise or column-wise traversal
• Thinking only diagonals are involved
• Ignoring the inward movement after outer layer

2. Which Rust loop structure is best suited to iterate over matrix indices for spiral traversal?

easy

A. loop { ... } with break condition

B. for i in 0..n { ... }

C. while i <= n { ... }

D. match statement for index matching

Solution

1. Step 1: Identify loop types in Rust

   Rust commonly uses for i in 0..n for fixed range iteration, ideal for matrix indices.

2. Step 2: Evaluate options for spiral traversal

   for i in 0..n { ... } uses range-based for loop, perfect for controlled index traversal. while i <= n { ... } is less idiomatic. loop { ... } with break condition is infinite loop needing manual breaks. match statement for index matching is for pattern matching, not iteration.

3. Final Answer:

   for i in 0..n { ... } -> Option B

4. Quick Check:

   Range-based for loop = best for index iteration [OK]

Hint: Use for loops with ranges for matrix indices [OK]

Common Mistakes:

• Using infinite loops without proper break
• Using while loops without clear index control
• Misusing match for iteration

3. What is the output of the spiral traversal on this 3x3 matrix?

let matrix = vec![
  vec![1, 2, 3],
  vec![4, 5, 6],
  vec![7, 8, 9],
];
// Spiral traversal code prints elements in order

medium

A. [1, 2, 3, 6, 9, 8, 7, 4, 5]

B. [1, 4, 7, 8, 9, 6, 3, 2, 5]

C. [1, 2, 3, 4, 5, 6, 7, 8, 9]

D. [9, 8, 7, 6, 5, 4, 3, 2, 1]

Solution

1. Step 1: Trace spiral traversal order

   Start top row left to right: 1, 2, 3; then right column down: 6, 9; bottom row right to left: 8, 7; left column up: 4; center element: 5.

2. Step 2: Write elements in order

   Combined order is [1, 2, 3, 6, 9, 8, 7, 4, 5].

3. Final Answer:

   [1, 2, 3, 6, 9, 8, 7, 4, 5] -> Option A

4. Quick Check:

   Outer to inner spiral = [1,2,3,6,9,8,7,4,5] [OK]

Hint: Follow edges clockwise, then move inward [OK]

Common Mistakes:

• Listing elements row-wise instead of spiral
• Reversing order of bottom or left edges
• Skipping center element

4. Identify the bug in this Rust spiral traversal snippet:

let mut top = 0;
let mut bottom = matrix.len() - 1;
let mut left = 0;
let mut right = matrix[0].len() - 1;
while top <= bottom && left <= right {
  for i in left..=right {
    print!("{} ", matrix[top][i]);
  }
  top += 1;
  for i in top..=bottom {
    print!("{} ", matrix[i][right]);
  }
  right -= 1;
  for i in (left..=right).rev() {
    print!("{} ", matrix[bottom][i]);
  }
  bottom -= 1;
  for i in (top..=bottom).rev() {
    print!("{} ", matrix[i][left]);
  }
  left += 1;
}

medium

A. The loop may print elements twice when top > bottom or left > right

B. The ranges in for loops are incorrect and cause syntax errors

C. The variables top, bottom, left, right are not initialized properly

D. The print statements use wrong matrix indices causing runtime panic

Solution

1. Step 1: Analyze loop conditions and updates

   The while loop condition is correct, but inside the loop, after updating top, bottom, left, right, the for loops still run without checking if top <= bottom or left <= right again.

2. Step 2: Identify consequence of missing checks

   This can cause the inner loops to run when top > bottom or left > right, printing elements twice or causing invalid access.

3. Final Answer:

   The loop may print elements twice when top > bottom or left > right -> Option A

4. Quick Check:

   Missing inner boundary checks = duplicate prints [OK]

Hint: Check boundaries before each edge traversal [OK]

Common Mistakes:

• Not checking boundaries inside the loop after updates
• Assuming while condition covers all cases
• Ignoring reverse ranges correctness

5. Given a rectangular matrix with dimensions 3x5, how many elements will be printed in spiral traversal, and what is the order of traversal directions?

hard

A. 15 elements; order: left to right, bottom to top, right to left, top to bottom, repeat

B. 15 elements; order: top to bottom, left to right, bottom to top, right to left, repeat

C. 8 elements; order: left to right, top to bottom, right to left, bottom to top, repeat

D. 15 elements; order: left to right, top to bottom, right to left, bottom to top, repeat

Solution

1. Step 1: Calculate total elements in 3x5 matrix

   3 rows * 5 columns = 15 elements to print in spiral traversal.

2. Step 2: Identify traversal direction order

   Spiral traversal moves: left to right (top row), top to bottom (right column), right to left (bottom row), bottom to top (left column), then repeats inward.

3. Final Answer:

   15 elements; order: left to right, top to bottom, right to left, bottom to top, repeat -> Option D

4. Quick Check:

   3*5=15; spiral directions = LR, TB, RL, BT [OK]

Hint: Count total elements; remember spiral direction cycle [OK]

Common Mistakes:

• Miscounting total elements
• Mixing up traversal direction order
• Assuming fewer elements printed